Question

A loop circuit has a resistance of R1 and a current of 2 A. The current is reduced to 1.5 A when an additional 1.6 Ω resistor is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.

Answer 1

Answer:

R1 = 4.8Ω

Explanation:

The loop circuit has an initial voltage of  V = IR

I = 2 A , R1 = R

V = 2R1

with the current reduced to 1.5A with an additional 1.6Ω resistor

the total resistance of the circuit is 1.6 + R1

the voltage of the two scenarios has to be equal , since the same voltage flows through the circuit

therefore V = 2R1

from Ohms law V = IR

2R1= 1.5 (1.6 + R1)

2R1 = 2.4 + 1.5R1

collecting like terms

2R1 - 1.5R1 = 2.4

0.5R1 = 2.4

R1 = $\frac{2.4}{0.5}$

R1 = 4.8Ω

Answer 2

Answer:

4.8 Ω

Explanation:

From Ohm's Law,

Using,

I = E/(R+r)................. Equation 1

E = I(R+r)................. Equation 2

Where I = current, E = emf, R = external resistance, r = internal resistance

Given: I = 2 A, R = R1, r = 0 Ω

Substitute into equation 2

E = 2(R1)

E = 2R1.

When an additional 1.6 Ω  resistor is added in series,

E = 1.5(R1+1.6)

2R1 = 1.5R1+2.4

2R1-1.5R1 = 2.4

0.5R1 = 2.4

R1 = 2.4/0.5

R1 = 4.8 Ω